# Goodness of Fit for Poisson Regression using R

I am trying to determine how well a Poisson model fits my data using Residual Null and Saturated Deviances. The Y col is the # of pennies that landed in a cup and the cup column represents the size. This is the R-code I have so far.

Y=c(4,7,2,5,4,5,10,6,8,2,9,9,9,7,10)
Cup=c(0,2,0,0,0,1,1,1,2,1,2,0,2,1,2)

sum(Y)
M1=glm(Y~Cup,family="poisson")
M1$deviance SaturatedProbs=dpois(Y,Y) SaturatedProbs Ls=prod(SaturatedProbs) Ds=-2*log(Ls) Ds #NullProbs ln=sum(Y)/length(Y) probsn=dpois(Y,ln) Ln=prod(probsn) Ln Dn=-2*log(Ln) Dn Dn-Ds  Ultimately, I want to compare my model to a chi_square distribution. I am having some trouble determining the appropriate inputs. Is what I have so far enough to determine the goodness-of-fit? ## 1 Answer You would need the deviances and the degrees of freedom to perform a deviance test. If the model fits the data well then$D_1 \sim \chi^2(n-p)$and$D_2 \sim \chi^2(n-q)$.$D_1$and$D_2$are deviances for model 1 and model 2.$n$is the number of parameters for saturated model.$p$and$q$are the number of parameters for given models ($q < p < n\$)

Deviance test:

$$D_1 - D_2 \sim \chi^2(p-q)$$

If value is greater than expected from chi squared, reject model 1.

• Okay so, please correct me if I am wrong but I am comparing my null model, which only contains 1 parameter, with my fitted model, which contains 2 parameters? The deviance for my null model is Dn=72, the deviance for the saturated model is Ds=54, and the residual deviance for the fitted model is Dr=12. I am looking at the drop in deviance of my null model to my fitted model. I have Dn-((Dn-Ds)-Dr) or 72-(17-12) = 66. Is this mathematically sound so far? How again does this look on the chi_sq distribution with 13 degrees of freedom? Do I simply compare this to my cut off at 0.05? Jun 4, 2018 at 23:22